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Starting in 4, we gave you a fully functioning web site with database persistence. In this chapter, we have expanded that relatively crude database, and now you essentially have a fully functioning schema complete with constraints and relations. Now that you have this wonderful database to use, it s time to start using it. We re going to assume that if you re reading this book, you have at least cursory knowledge of creating SQL queries. However, if that s not the case, don t worry. Grails has made creating SQL queries extremely easy in fact, it s almost too easy, if you ask us. To begin, we ll go over four different ways to query the database. This might seem like an overly excessive way of querying the database, but in the end, you will find that you have the flexibility to create a query with as little or as much information as you need to provide. We ll show you how to do simple CRUD operations. We demonstrated this in the previous two chapters, but we ll briefly rehash it so you can see how to build up your querying ability. We ll then explain how GORM really shows off its DSL capabilities by being able to create dynamic queries in the form of method names. You saw a bit of this in the earlier chapters when we did findBy, but now we ll show you all the options and parameters available to you. We ll cover this for both straight single retrievals and for retrieving multiple objects. Finally, we ll show you how to use Hibernate Query Language (HQL) queries instead of the more dynamic DSL queries. Sometimes using HQL is the only way to get the query you want.

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Attaching to a process that belongs to a RAC instance on another node in the same cluster is not implemented. The commands, their arguments, and how to obtain argument values are summarized in Table 37-1.

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vb.net pdf 417 reader

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7-12 We need to show this for the general case where we have a set of edges that we know are going into the solution The subproblem is then the remaining graph, and we want to show that finding a minimum spanning tree in the rest that s compatible with what we have (no cycles) will give us an optimal solution globally As usual, we show this by contradiction, assuming that we could find a nonoptimal solution to this subproblem that would give us a better global solution Both subsolutions would be compatible with what we had, so they would be interchangeable Clearly, swapping out the nonoptimal solution with the optimal one would improve the global sum, which gives us our contradiction 7-13 Kruskal s algorithm invariably finds a minimum spanning forest, which in the case of connected graphs turns into a minimum spanning tree.

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When interacting with a database, you need to know how to do basic CRUD operations. Most Java developers reading this are probably used to the standard DAO domain model, where after you create the domain, you create a DAO with various operations. These DAO models usually have the standard void delete(Domain d) and get(String primaryKey) methods. Before Hibernate, these methods would usually interact with the database directly with straight SQL calls. This kind of methodology made a bit of sense when you had to write the SQL code directly, but with today s tools and a dynamic language like Groovy, these constraints are no longer necessary. Hibernate helped eliminate the problem of having to hard-code SQL by allowing you to use more abstract terminology to persist to the database. But that solved only half the

Attach to your own server process Attach to a foreign process, where pid equals V$PROCESS.PID Attach to a foreign process, where spid equals V$PROCESS.SPID

Prim s algorithm could be extended with a loop, like depth first search, so that it would restart in all components 7-14 It will still run, but it won t necessarily find the cheapest traversal (or min-cost arborescence) 7-15 Because you can use this to sort real numbers, which has a loglinear lower bound (This is similar to the case with convex hulls) You just use the numbers as x-coordinates and use identical y-coordinates The minimum spanning tree would then be a path from the first number to the last, giving you the sorted order 7-16 All we need to show is that the component trees have (at most) logarithmic height The height of a component tree is equal to the highest rank in the component This rank is increased only if two component tree of equal height are merged, and then it is increased by one.

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